Solve for $x$ : $ 6|x - 1| + 2 = 3|x - 1| + 10 $
Explanation: Subtract $ {3|x - 1|} $ from both sides: $ \begin{eqnarray} 6|x - 1| + 2 &=& 3|x - 1| + 10 \\ \\ { - 3|x - 1|} && { - 3|x - 1|} \\ \\ 3|x - 1| + 2 &=& 10 \end{eqnarray} $ Subtract ${2}$ from both sides: $ \begin{eqnarray} 3|x - 1| + 2 &=& 10 \\ \\ { - 2} &=& { - 2} \\ \\ 3|x - 1| &=& 8 \end{eqnarray} $ Divide both sides by ${3}$ $ \dfrac{3|x - 1|} {{3}} = \dfrac{8} {{3}} $ Simplify: $ |x - 1| = \dfrac{8}{3}$ Because the absolute value of an expression is its distance from zero, it has two solutions, one negative and one positive: $ x - 1 = -\dfrac{8}{3} $ or $ x - 1 = \dfrac{8}{3} $ Solve for the solution where $x - 1$ is negative: $ x - 1 = -\dfrac{8}{3} $ Add ${1}$ to both sides: $ \begin{eqnarray} x - 1 &=& -\dfrac{8}{3} \\ \\ {+ 1} && {+ 1} \\ \\ x &=& -\dfrac{8}{3} + 1 \end{eqnarray} $ Change the ${ + 1}$ to an equivalent fraction with a denominator of $3$ $ x = - \dfrac{8}{3} {+ \dfrac{3}{3}} $ $ x = -\dfrac{5}{3} $ Then calculate the solution where $x - 1$ is positive: $ x - 1 = \dfrac{8}{3} $ Add ${1}$ to both sides: $ \begin{eqnarray} x - 1 &=& \dfrac{8}{3} \\ \\ {+ 1} && {+ 1} \\ \\ x &=& \dfrac{8}{3} + 1 \end{eqnarray} $ Change the ${ + 1}$ to an equivalent fraction with a denominator of $3$ $ x = \dfrac{8}{3} {+ \dfrac{3}{3}} $ $ x = \dfrac{11}{3} $ Thus, the correct answer is $x = -\dfrac{5}{3} $ or $x = \dfrac{11}{3} $.